Question

You hold a bucket in one hand. In the bucket is a 500 g rock. You swing the bucket so the rock moves in a vertical circle 2.2 m in diameter. What is the minimum speed the rock must have at the top of the circle if it to always stay-in contact with the bottom of the bucket?

Answer 1

Answer:v=3.28 m/s

Step-by-step explanation:

Given

mass of rock
`m=500 gm`

diameter of circle
`d=2.2 m`

radius
`r=(2.2)/(2)=1.1 m`

At highest Point

`mg+N=(mv^2)/(r)`

At highest Point N=0 because mass is just balanced by centripetal Force

thus
`mg=(mv^2)/(r)`

`v=√(gr)`

`v=√(9.8* 1.1)`

`v=√(10.78)`

`v=3.28 m/s`